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I am reading Perelman's paper "Elements of Morse theory on Aleksandrov spaces", St. Petersburg Math. J. 5 (1994), no. 1, 205–213. Here is a Russian version (I cold not find the English one).

There is a part of the proof I do not understand.

In the proof of Theorem 1.4 there is an induction base case $k=n$ which I do not understand. One has an $n$-dimensional Alexandrov space $M^n$, and an admissible map $g\colon M\to \mathbb{R}^n$.

It is claimed that in this case Theorem 1.4 follows from property 1.2. To make it more explicit here, property 1.2 says that any admissible map is Lipschitz and open near its regular point. The claim is that that implies in particular that an admissible map has a trivialization in a neighborhood of its regular point. WHY?

It seems that existence of trivialization in our situation is equivalent that the map $g$ is locally homeomorphism. That does not follow merely from the Lipschitz property and openness, and something else is necessary.

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    $\begingroup$ The base case is actually $k=n+1$, not $k=n$. In this case it is trivial because there are no regular points. $\endgroup$
    – John Harvey
    Commented Dec 13, 2022 at 11:16

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Indeed, openness does not solely imply regularity. (Say, double branching covering of the plane is open, but not regular.)

But the map is admissible and that is the key word. The definition of admissible map is bit technical, but main example is a map with distance functions as coordinates. In this case, openness yields regularity.

(Hope it helps.)

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